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[tex]A = \begin{bmatrix}1&2\\1&1\end{bmatrix} \implies A^{-1} = \dfrac1{\det(A)}\begin{bmatrix}1&-1\\-2&1\end{bmatrix} = \begin{bmatrix}-1&1\\2&-1\end{bmatrix}[/tex]
where det(A) = 1×1 - 2×1 = -1.
[tex]B = \begin{bmatrix}0&-1\\1&2\end{bmatrix} \implies B^{-1} = \dfrac1{\det(B)}\begin{bmatrix}2&1\\-1&0\end{bmatrix} = \begin{bmatrix}2&1\\-1&0\end{bmatrix}[/tex]
where det(B) = 0×2 - (-1)×1 = 1. Then
[tex]B^{-1}A^{-1} = \begin{bmatrix}2&1\\-1&0\end{bmatrix} \begin{bmatrix}-1&1\\2&-1\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}[/tex]
On the other side, we have
[tex]AB = \begin{bmatrix}1&2\\1&1\end{bmatrix} \begin{bmatrix}0&-1\\1&2\end{bmatrix} = \begin{bmatrix}2&3\\1&1\end{bmatrix}[/tex]
and det(AB) = det(A) det(B) = (-1)×1 = -1. So
[tex](AB)^{-1} = \dfrac1{\det(AB)}\begin{bmatrix}1&-3\\-1&2\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}[/tex]
and both matrices are clearly the same.
More generally, we have by definition of inverse,
[tex](AB)(AB)^{-1} = I[/tex]
where [tex]I[/tex] is the identity matrix. Multiply on the left by A ⁻¹ to get
[tex]A^{-1}(AB)(AB)^{-1} = A^{-1}I = A^{-1}[/tex]
Multiplication of matrices is associative, so we can regroup terms as
[tex](A^{-1}A)B(AB)^{-1} = A^{-1} \\\\ B(AB)^{-1} = A^{-1}[/tex]
Now multiply again on the left by B ⁻¹ and do the same thing:
[tex]B^{-1}\left(B(AB)^{-1}\right) = (B^{-1}B)(AB)^{-1} = B^{-1}A^{-1} \\\\ (AB)^{-1} = B^{-1}A^{-1}[/tex]